Suppose a Bézier curve is translated to \({\bf{x}}\left( t \right) + {\bf{b}}\) . That is, for\(0 \le t \le 1\), the new curve is

\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^3}{{\bf{p}}_o} + 3t{\left( {1 - t} \right)^2}{{\bf{p}}_1} + 3{t^2}\left( {1 - t} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3} + {\bf{b}}\)

Show that this new curve is again a Bézier curve.

If the curve is shifted to\({\rm{x}}\left( t \right) + {\rm{b}}\),then the new control points become\({{\rm{p}}_o} + {\rm{b}},\,\,{{\rm{p}}_1} + {\rm{b}},\,\,{{\rm{p}}_2} + {\rm{b}},\,\,{{\rm{p}}_3} + {\rm{b}}\).

The new Bezier curve is represented as shown below:

\(\begin{array}{c}{\bf{y}}\left( t \right) = {\left( {1 - t} \right)^3}\left( {{{\bf{p}}_o} + {\bf{b}}} \right) + 3t\left( {1 - {t^2}} \right)\left( {{{\bf{p}}_1} + {\bf{b}}} \right) + 3{t^2}\left( {1 - t} \right)\left( {{{\bf{p}}_2} + {\bf{b}}} \right) + {t^3}\left( {{{\bf{p}}_3} + {\bf{b}}} \right)\\ = {\left( {1 - t} \right)^3}{{\bf{p}}_o} + 3t\left( {1 - {t^2}} \right){{\bf{p}}_1} + 3{t^2}\left( {1 - t} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3} + {\left( {1 - t} \right)^3}{\bf{b}} + 3t\left( {1 - {t^2}} \right){\bf{b}}\\ + 3{t^2}\left( {1 - t} \right){\bf{b}} + {t^3}{\bf{b}}\end{array}\)

The weighted sum of linear combination must be equal to 1.

\(\begin{array}{c}{\left( {1 - t} \right)^3} + 3t\left( {1 - {t^2}} \right) + 3{t^2}\left( {1 - t} \right) + {t^3} = 1 - {t^3} - 3t + 3{t^2} + 3t - 3{t^2} + {t^3}\\ = 1\end{array}\)

As \({\left( {1 - t} \right)^3} + 3t\left( {1 - {t^2}} \right) + 3{t^2}\left( {1 - t} \right) + {t^3} = 1\), for all \(t\). This implies that \({\rm{y}}\left( t \right) = {\rm{x}}\left( t \right) + {\rm{b}}\) is also true for all \(t\).

Thus, translation by \(b\)maps a Bézier curve into a new Bézier curve.